Question 14 Marks
Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf $\in.$ All surfaces are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.
Answer
We knows
In this particular case the electricfield attracts the dielectric into the capacitor with a force $\frac{\in_0\text{bV}^2(\text{k}-1)}{2\text{d}}$
Where, b = Width of plates
k = Dielectric constant
d = Separation between plates
V = E = Potential difference.
Hence in this case the surfaces are frictionless, this force is counteracted by the weight.
So, $\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{d}}=\text{Mg}$
$\Rightarrow\text{M}=\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{dg}}$
View full question & answer→We knows
In this particular case the electricfield attracts the dielectric into the capacitor with a force $\frac{\in_0\text{bV}^2(\text{k}-1)}{2\text{d}}$
Where, b = Width of plates
k = Dielectric constant
d = Separation between plates
V = E = Potential difference.
Hence in this case the surfaces are frictionless, this force is counteracted by the weight.
So, $\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{d}}=\text{Mg}$
$\Rightarrow\text{M}=\frac{\in_0\text{bE}^2(\text{k}-1)}{2\text{dg}}$